Algorithms for deletion problems on split graphs

TL;DR

This paper develops fixed-parameter algorithms for split graph deletion problems, specifically transforming split graphs into block or threshold graphs efficiently with exponential algorithms.

Contribution

The paper introduces new algorithms with specific exponential running times for Split to Block Vertex Deletion and Split to Threshold Vertex Deletion problems.

Findings

Algorithms run in $O^*(2.076^k)$ and $O^*(2.733^k)$ time

Efficient fixed-parameter algorithms for split graph deletion problems

Improved computational methods for graph modification problems

Abstract

In the Split to Block Vertex Deletion and Split to Threshold Vertex Deletion problems the input is a split graph $G$ and an integer $k$, and the goal is to decide whether there is a set $S$ of at most $k$ vertices such that $G-S$ is a block graph and $G-S$ is a threshold graph, respectively. In this paper we give algorithms for these problems whose running times are $O^*(2.076^k)$ and $O^*(2.733^k)$, respectively.

Equations

$$T(k)\le T(k-c_1)+\cdots +T(k-c_t).$$

Full text

Keywords

graph algorithms, parameterized complexity.

1 Introduction

A graph $G$ is called a split graph if its vertex set can be partitioned into two disjoint sets $C$ and $I$ such that $C$ is a clique and $I$ is an independent set. A graph $G$ is a block graph if every biconnected component of $G$ is a clique. A graph $G$ is a threshold graph if there is a $t\in\mathbb{R}$ and a function $f\colon V(G)\to\mathbb{R}$ such that for every $u,v\in V(G)$, $(u,v)$ is an edge in $G$ if and only if $f(u)+f(v)\geq t$.

In the Split to Block Vertex Deletion (SBVD) problem the input is a split graph $G$ and an integer $k$, and the goal is to decide whether there is a set $S$ of at most $k$ vertices such that $G-S$ is a block graph. Similarly, in the Split to Threshold Vertex Deletion (STVD) problem the input is a split graph $G$ and an integer $k$, and the goal is to decide whether there is a set $S$ of at most $k$ vertices such that $G-S$ is a threshold graph. The SBVD and STVD problems were shown to be NP-hard by Cao et al. [1]. A split graph $G$ is a block graph if and only if $G$ does not contain an induced diamond, where a diamond is a graph with $4$ vertices and $5$ edges. Additionally, a split graph $G$ is threshold graph if and only if $G$ does not contain an induced path with 4 vertices. Therefore, SBVD and STVD are special cases of the 4-Hitting Set problem. Using the fastest known parameterized algorithm for 4-Hitting Set, due to Fomin et al. [4], the SBVD and STVD problems can be solved in $O^{*}(3.076^{k})$ time. Choudhary et al. [2] gave faster algorithms for SBVD and STVD whose running times are $O^{*}(2.303^{k})$ and $O^{*}(2.792^{k})$, respectively. In this paper we give algorithms for SBVD and STVD whose running times are $O^{*}(2.076^{k})$ and $O^{*}(2.733^{k})$, respectively.

2 Preliminaries

For a graph $G$ and a vertex $v\in V(G)$, $N(v)$ is the set of vertices that are adjacent to $v$. For a set $S$ of vertices, $G-S$ is the graph obtained from $G$ by deleting the vertices of $S$ (and incident edges). Let $P_{4}$ denote a graph that is a path on 4 vertices.

In the 3-Hitting Set problem the input is a family $\mathcal{F}$ of subsets of size at most 3 of a set $U$ and an integer $k$, and the goal is to decide whether there is a set $X\subseteq U$ of size at most $k$ such that $X\cap A\neq\emptyset$ for every $A\in\mathcal{F}$.

For two families of sets $\mathcal{A}$ and $\mathcal{B}$, $\mathcal{A}\circ\mathcal{B}=\{A\cup B\colon A\in\mathcal{A},B\in\mathcal{B}\}$.

2.1 Branching algorithm

A branching algorithm (cf. [3]) for a parameterized problem is a recursive algorithm that uses rules. Given an instance $(G,k)$ to the problem, the algorithm applies some rule. In each rule, the algorithm either computes the answer to the instance $(G,k)$, or performs recursive calls on instances $(G_{1},k-c_{1}),\ldots,(G_{t},k-c_{t})$, where $c_{1},\ldots,c_{t}>0$. The algorithm returns ‘yes’ if and only if at least one recursive call returned ‘yes’. The rule is called a reduction rule if $t=1$, and a branching rule if $t\geq 2$. To analyze the time complexity of the algorithm, define $T(k)$ to be the maximum number of leaves in the recursion tree of the algorithm when the algorithm is run on an instance with parameter $k$. Each branching rule corresponds to a recurrence on $T(k)$:

[TABLE]

The largest real root of $P(x)=1-\sum_{i=1}^{t}x^{-c_{i}}$ is called the branching number of the rule. The vector $(c_{1},\ldots,c_{t})$ is called the branching vector of the rule.

Let $\gamma$ be the maximum branching number over all branching rules. Assuming that the application of a rule takes $O^{*}(1)$ time, the time complexity of the algorithm is $O^{*}(\gamma^{k})$.

3 Algorithm for SBVD

Lemma 1**.**

Let $G$ be a split graph with a partition $C,I$ of its vertices. $G$ is a block graph if and only if (1) A vertex in $I$ with degree at least $2$ is adjacent to all vertices in $C$, and (2) There is at most one vertex in $I$ with degree at least 2.

Proof.

Suppose that $G$ is a block graph. If a vertex $v\in I$ has degree at least 2, let $a_{1},a_{2}\in C$ be two neighbors of $v$. For every $b\in C\setminus\{a_{1},a_{2}\}$, we have that $v$ is adjacent o $b$, otherwise $v,a_{1},a_{2},b$ induces a diamond, contradicting the assumption that $G$ is a block graph.

Now, suppose conversely that there are $u,v\in I$ with degree at least 2. Let $a_{1},a_{2}$ be two vertices in $C$. From the paragraph above, $a_{1},a_{2}$ are neighbors of $u$ and of $v$. Therefore, $u,v,a_{1},a_{2}$ induces a diamond, contradicting the assumption that $G$ is a block graph.

To prove the opposite direction, suppose that $G$ satisfied (1) and (2). Suppose conversely that $G$ is not a block graph. Then there is a set of vertices $X$ that induces a diamond. Since $C$ is a clique and $I$ is an independent set, $|X\cap I|$ is equal to either 1 or 2. If $|X\cap I|=1$ then $G$ does not satisfy (1), and if $|X\cap I|=2$ then $G$ does not satisfy (2), a contradiction. Therefore $G$ is a block graph. ∎

IF $(G,k)$ is a yes instance, let $S$ be a solution for $(G,k)$. By Lemma 1, there is at most one vertex in $I\setminus S$ with degree at least 2 in $G-S$. Denote this vertex, if it exists, by $v^{*}$. For every $v\in I\setminus(\{v^{*}\}\cup S)$ we have that $v$ has degree at most 1 in $G-S$.

The algorithm for SBVD goes over all possible choices for the vertex $v^{*}\in I$. Additionally, the algorithm also inspects the case in which no such vertex exist. For every choice of $v^{*}$, the algorithm deletes from $G$ all the vertices of $C$ that are not adjacent to $v^{*}$ and decreases the value of $k$ by the number of vertices deleted. When the algorithm inspects the case when $v^{*}$ does not exists, the graph is not modified.

For every choice of $v^{*}$, the algorithm generates an instance $(\mathcal{F},k)$ of 3-Hitting Set as follows: For every vertex $v\in I\setminus\{v^{*}\}$ that has at least two neighbors, and for every two neighbors $a,b\in C$ of $v$, the algorithm adds the set $\{v,a,b\}$ to $\mathcal{F}$. The algorithm then uses the algorithm of Wahlström [5] to solve the instance $(\mathcal{F},k)$ in $O^{*}(2.076^{k})$ time. If $(\mathcal{F},k)$ is a yes instance of 3-Hitting Set then the algorithm returns yes. If all the constructed 3-Hitting Set instances, for all choices of $v^{*}$, are no instances, the algorithm returns no.

4 Algorithm for STVD

Let $I_{0}$ be the set of all vertices in $I$ that have minimum degree (namely, a vertex $u\in I$ is in $I_{0}$ if $|N(u)|\leq|N(v)|$ for every $v\in I$). We say that two vertices $u,v\in I$ are twins if $N(u)=N(v)$. Let $\mathrm{Twins}(v)$ be a set containing $v$ and all the twins of $v$. Recall that a split graph $G$ is a threshold graph if and only if $G$ does not contain an induced $P_{4}$. Note that an induced $P_{4}$ in $G$ must be of the form $u,a,b,v$ where $u,v\in I$ and $a,b\in C$.

The algorithm for STVD is a branching algorithm. At each step, the algorithm applies the first applicable rule from the rules below. The reduction rules of the algorithm are as follows.

(R1)  If $k\leq 0$ and $G$ is not a threshold graph, return ‘no’.

(R2)  If $G$ is an empty graph, return ‘yes’.

(R3)  If $v$ is a vertex such that there is no induced $P_{4}$ in $G$ that contains $v$, delete $v$.

If Rule (R4) cannot be applied we have that for every $a\in C$ there is a vertex $v\in I$ such that $a\notin N(v)$.

We now describe the branching rules of the algorithm. When we say that the algorithm branches on sets $S_{1},\ldots,S_{p}$, we mean that the algorithm is called recursively on the instances $(G-S_{1},k-|S_{1}|),\ldots,(G-S_{p},k-|S_{p}|)$.

(B1)  If there are non-twin vertices $u,v\in I$ such that $|N(u)|=|N(v)|=1$, branch on $N(u)$ and $N(v)$.

To show the safeness of Rule (B1), denote $N(u)=\{a\}$ and $N(v)=\{b\}$. If $S$ is a solution for the instance $(G,k)$ then $S$ must contain at least one vertex from the induced path $u,a,b,v$. If $u\in S$ then $S^{\prime}=(S\setminus\{u\})\cup\{a\}$ is also a solution (since every induced $P_{4}$ that contains $u$ also contains $a$). Additionally, if $v\in S$ then $(S\setminus\{v\})\cup\{b\}$ is also a solution. Therefore, there is a solution $S$ such that either $a\in S$ or $b\in S$. Thus, Rule (B1) is safe.

The branching vector of Rule (B1) is $(1,1)$.

(B2)  If there is a vertex $u\in I$ such that $|N(u)|=1$, let $v\in I$ be a vertex such that $N(u)\not\subseteq N(v)$. Branch on $\{v\}$, $N(u)$ and $N(v)$.

Note that the vertex $v$ exists since Rule (R4) cannot be applied. To prove the safeness of Rule (B2), note that if $S$ is a solution for the instance $(G,k)$ then either $u\in S$, $v\in S$, $N(u)\subseteq S$, or $N(v)\subseteq S$. If one of the last three cases occurs we are done. Otherwise (if $u\in S$), $S^{\prime}=(S\setminus\{u\})\cup N(u)$ is also a solution. It follows that Rule (B2) is safe.

Since Rule (B4) cannot be applied, $|N(v)|\geq 2$. Therefore, the branching vector of Rule (B2) is at least $(1,1,2)$.

Note that if Rule (B2) cannot be applied, every vertex in $I$ has degree at least 2.

(B3)  If there are vertices $u,v\in I$ such that $|N(u)\setminus N(v)|\geq 2$ and $|N(v)\setminus N(u)|\geq 2$, branch on $\{u\}$, $\{v\}$, $N(u)\setminus N(v)$, and $N(v)\setminus N(u)$.

If $S$ is a solution for the instance $(G,k)$ then either $u\in S$, $v\in S$, $N(u)\setminus N(v)\subseteq S$, or $N(v)\setminus N(u)\subseteq S$ (If neither of the above cases hold, let $a\in(N(u)\setminus N(v))\setminus S$ and $b\in(N(v)\setminus N(u))\setminus S$. Then, $u,a,b,v$ is an induced $P_{4}$ in $G-S$, a contradiction). Therefore, Rule (B3) is safe.

The branching vector of Rule (B3) is at least $(1,1,2,2)$.

Lemma 2**.**

If Rule (B3) cannot be applied and $u,v\in I$ are two vertices such that $|N(u)|\leq|N(v)|$ then $|N(u)\setminus N(v)|\leq 1$.

Proof.

Suppose conversely that $|N(u)\setminus N(v)|\geq 2$. Then, $|N(v)\setminus N(u)|\geq|N(u)\setminus N(v)|\geq 2$. Therefore, Rule (B3) can be applied on $u,v$, a contradiction. ∎

We now consider two cases.

Case 1

In the first case, every two vertices in $I_{0}$ are twins. The algorithm picks an arbitrary vertex $u\in I_{0}$ and vertices $a_{1},a_{2}\in N(u)$. Since Rule (R4) cannot be applied, there is a vertex $v_{1}\in I$ such that $a_{1}\notin N(v_{1})$ and a vertex $v_{2}\in I$ such that $a_{2}\notin N(v_{2})$. For $i=1,2$ we have that $v_{i}\notin I_{0}$ since $v_{i}$ is not a twin of $u$. Since $u\in I_{0}$, it follows that $|N(v_{i})|>|N(u)|$. Thus, $|N(v_{i})\setminus N(u)|>|N(u)\setminus N(v_{i})|\geq 1$. By Lemma 2 and the fact that $a_{i}\in N(u)\setminus N(v_{i})$ we obtain that $N(u)\setminus N(v_{i})=\{a_{i}\}$ and thus $N(u)\setminus\{a_{i}\}\subseteq N(v_{i})$. In particular, $a_{2}\in N(v_{1})$ and $a_{1}\in N(v_{2})$. Note that this implies that $v_{1}\neq v_{2}$.

Lemma 3**.**

$|(N(v_{1})\cap N(v_{2}))\setminus N(u)|\geq 2$.

Proof.

Suppose without loss of generality that $|N(v_{1})|\leq|N(v_{2})|$. By Lemma 2 and the fact that $a_{2}\in N(v_{1})\setminus N(v_{2})$ we have that $N(v_{1})\setminus N(u)\subseteq N(v_{2})$. Therefore, $(N(v_{1})\cap N(v_{2}))\setminus N(u)=N(v_{2})\setminus N(u)$. We have shown above that $|N(v_{2})\setminus N(u)|\geq 2$. ∎

(B4)  If Case 1 occurs and $a_{1},a_{2}\in N(w)$ for every $w\in I\setminus\{v_{1},v_{2}\}$, branch on $\{u\}$, $(N(v_{1})\cap N(v_{2}))\setminus N(u)$, and $\{v_{1},v_{2}\}$.

We now prove the safeness of Rule (B4). In order to delete the paths of the form $u,a_{1},b,v_{1}$ or $u,a_{2},b,v_{2}$ for some $b\in(N(v_{1})\cap N(v_{2}))\setminus N(u)$, a solution $S$ must satisfy one of the following (1) $u\in S$ (2) $(N(v_{1})\cap N(v_{2}))\setminus N(u)\subseteq S$, or (3) $S$ contains at least one vertex from $\{a_{1},v_{1}\}$ and at least one vertex from $\{a_{2},v_{2}\}$. Suppose that $S$ is a solution that satisfies (3). Due to the assumption of Rule (B4) and the fact that $a_{1}\in N(v_{2})$, we have that every vertex in $I\setminus\{v_{1}\}$ is adjacent to $a_{1}$. Therefore, every induced $P_{4}$ that contains $a_{1}$ is of the form $v_{1},x,a_{1},y$. Thus, if $v_{1}\notin S$ then $S^{\prime}=(S\setminus\{a_{1}\})\cup\{v_{1}\}$ is also a solution. Similarly, if $v_{2}\notin S$ then $S^{\prime}=(S\setminus\{a_{2}\})\cup\{v_{2}\}$ is also a solution. Therefore, if $(G,k)$ is a yes instance, there is a solution $S$ such that either $S$ satisfies (1) or (2) above, or $\{v_{1},v_{2}\}\subseteq S$.

By Lemma 3, the branching vector of Rule (B4) is at least $(1,2,2)$.

(B5)  If Case 1 occurs, let $w\in I\setminus\{v_{1},v_{2}\}$ be a vertex such that $\{a_{1},a_{2}\}\not\subseteq N(w)$, and without loss of generality assume that $w_{1}\notin N(w)$. Branch on $\{u\}$, $(N(v_{1})\cap N(v_{2}))\setminus N(u)$, and on the sets in $\{\{a_{1}\},\{v_{1}\}\cup(N(w)\setminus N(u)),\{v_{1},w\}\}\circ\{\{a_{2}\},\{v_{2}\}\}$.

We now show the safeness of Rule (B5). In order to delete the induced paths of the form $u,a_{1},b,v_{1}$ or $u,a_{2},b,v_{2}$ for $b\in(N(v_{1})\cap N(v_{2}))\setminus N(u)$, a solution $S$ must satisfy (1), (2), or (3) above. Suppose that (1) is not satisfied (namely, $u\notin S$) and that (3) is satisfied. Additionally, suppose that $a_{1}\notin S$. Therefore, $v_{1}\in S$ and $S$ contains at least one vertex from $\{a_{2},v_{2}\}$. In order to delete the induced paths of the form $u,a_{1},c,w$ for every $c\in N(w)\setminus N(u)$, either $w\in S$ or $N(w)\setminus N(u)\subseteq S$. We have that $w\notin I_{0}$ since $w$ is not a twin of $u$. Since $u\in I_{0}$, it follows that $|N(w)|>|N(u)|$. Thus, $|N(w)\setminus N(u)|>|N(u)\setminus N(w)|\geq 1$. From the previous inequality, Lemma 3, and the fact that $v_{1},v_{2},a_{2}\notin N(w)\setminus N(u)$, it follows that the branching vector of Rule (B5) is at least $(1,2,2,4,3,2,4,3)$.

Case 2

In the second case, there are non-twin vertices in $I_{0}$. Suppose that $u_{1},u_{2}\in I_{0}$ are non-twin vertices, where the choice of $u_{1},u_{2}$ will be given later. By Lemma 2, $|N(u_{1})\setminus N(u_{2})|=|N(u_{2})\setminus N(u_{1})|=1$. Denote $N(u_{1})\setminus N(u_{2})=\{a_{1}\}$ and $N(u_{2})\setminus N(u_{1})=\{a_{2}\}$. Let $I_{1}=I_{0}\setminus(\mathrm{Twins}(u_{1})\cup\mathrm{Twins}(u_{2}))$.

Lemma 4**.**

If $I_{1}\neq\emptyset$ then either (1) for every $u\in I_{1}$, $N(u)$ consists of $N(u_{1})\cap N(u_{2})$ plus an additional vertex that is not in $\{a_{1},a_{2}\}$, or (2) for every $u\in I_{1}$, $N(u)$ consists of $a_{1}$, $a_{2}$, and all the vertices of $N(u_{1})\cap N(u_{2})$ except one vertex.

Proof.

We first claim that every $u\in I_{1}$, $|N(u)\cap{a_{1},a_{2}}|$ is either 0 or 2. Suppose conversely that $N(u)$ contains exactly one vertex from $a_{1},a_{2}$ and without loss of generality, $a_{2}\in N(u)$ and $a_{1}\notin N(u)$. By Lemma 2 on $u,u_{1}$ we obtain that $N(u)$ contains all the vertices in $N(u_{1})\setminus\{a_{1}\}=N(u_{2})\setminus\{a_{2}\}$. Since we assumed that $a_{2}\in N(u)$, we have that $N(u_{2})\subseteq N(u)$. From the fact that $|N(u_{2})|=|N(u)|$ we obtain that $N(u_{2})=N(u)$, contradicting the assumption that $u$ is not a twin of $u_{2}$. Therefore, $|N(u)\cap{a_{1},a_{2}}|$ is either 0 or 2.

We first assume that there is no vertex $u\in I_{1}$ such that $a_{1},a_{2}\in N(u)$. From the claim above we have that $a_{1}\notin N(u)$. By Lemma 2 on $u,u_{1}$, $N(u)$ contains all the vertices in $N(u_{1})\setminus\{a_{1}\}=N(u_{1})\cap N(u_{2})$ plus an additional vertex that is not in $\{a_{1},a_{2}\}$.

Now suppose that there is a vertex $u_{3}\in I_{1}$ such that $a_{1},a_{2}\in N(u_{3})$. Consider some $u\in I_{1}$. We claim that $a_{1}\in N(u)$. Suppose conversely that $a_{1}\notin N(u)$. From the claim above, $a_{2}\notin N(u)$. Therefore, $a_{1},a_{2}\in N(u_{3})\setminus N(u)$, contradicting Lemma 2. Thus, $a_{1}\in N(u)$. From the claim above and Lemma 2 we conclude that $N(u)$ contains $a_{2}$ and all the vertices in $N(u_{1})\cap N(u_{2})$ except one vertex. ∎

If $I_{1}=\emptyset$ or the first case of Lemma 4 occurs, we say that the vertices of $I_{0}$ form a sunflower. Note that if the vertices of $I_{0}$ do not form a sunflower, for every vertex $a\in N(u_{1})\cup N(u_{2})$ there are non-twin vertices $u,u^{\prime}\in I_{0}$ that are adjacent to $a$.

(B6)  If there are non-twin vertices $u_{1},u_{2}\in I_{0}$ such that $a_{1},a_{2}\in N(w)$ for every $w\in I\setminus I_{0}$, then suppose without loss of generality that $|\mathrm{Twins}(u_{1})|\leq|\mathrm{Twins}(u_{2})|$. Branch on $\mathrm{Twins}(u_{1})$ and $\{a_{1}\}$.

To prove the safeness of Rule (B6), suppose that $(G,k)$ is a yes instance and let $S$ be a solution. If $a_{1}\in S$ or $\mathrm{Twins}(u_{1})\subseteq S$ we are done, so suppose that that $a_{1}\notin S$ and $\mathrm{Twins}(u_{1})\not\subseteq S$. We can assume that $S\cap\mathrm{Twins}(u_{1})=\emptyset$ (otherwise, $S^{\prime}=S\setminus\mathrm{Twins}(u_{1})$ is also a solution). Since $u^{\prime}_{1},a_{1},a_{2},u^{\prime}_{2}$ is an induced path for every $u^{\prime}_{1}\in\mathrm{Twins}(u_{1})$ and $u^{\prime}_{2}\in\mathrm{Twins}(u_{2})$, either $a_{2}\in S$ or $\mathrm{Twins}(u_{2})\subseteq S$. Note that we can assume that if $S$ contains at least one vertex from $\mathrm{Twins}(u_{2})$ then it contains all the vertices of $\mathrm{Twins}(u_{2})$. Define a set $S^{\prime}$ by taking the vertices in $S\setminus(\{a_{2}\}\cup\mathrm{Twins}(u_{2}))$. Additionally, if $a_{2}\in S$, add $a_{1}$ to $S^{\prime}$, and if $\mathrm{Twins}(u_{2})\subseteq S$, add $\mathrm{Twins}(u_{1})$ to $S^{\prime}$. We now show that $S^{\prime}$ is also a solution. Since $|\mathrm{Twins}(u_{1})|\leq|\mathrm{Twins}(u_{2})|$, we have that $|S^{\prime}|\leq|S|\leq k$. Suppose conversely that $G-S^{\prime}$ contains an induced $P_{4}$ and denote this path by $P^{\prime}$. Create a path $P$ by taking $P^{\prime}$ and performing the following steps: (1) If $a_{1}$ is in $P^{\prime}$, replace it with $a_{2}$. (2) If $a_{2}$ is in $P^{\prime}$, replace it with $a_{1}$. (3) If $P^{\prime}$ contains a vertex $u^{\prime}_{1}\in\mathrm{Twins}(u_{1})$, replace it with $u_{2}$. (4) If $P^{\prime}$ contains a vertex $u^{\prime}_{2}\in\mathrm{Twins}(u_{2})$, replace it with $u_{1}$. Recall that $a_{1},a_{2}\in N(w)$ for every $w\in I\setminus I_{0}$. Additionally, for every $u\in I_{1}$, $a_{1},a_{2}\notin N(u)$ if the vertices of $I_{0}$ form a sunflower, and $a_{1},a_{2}\in N(u)$ otherwise. Therefore, for every two vertices $x^{\prime},y^{\prime}$ in $P^{\prime}$ and the corresponding vertices $x,y$ in $P$, we have that $(x,y)$ is an edge if and only if $(x^{\prime},y^{\prime})$ is an edge. It follows that $P$ is also an induced path in $G$. From the assumptions that $a_{1}\notin S$ and $\mathrm{Twins}(u_{1})\cap S=\emptyset$ and from the definition of $S^{\prime}$ we have that $S$ does not contain a vertex of $P$. This contradicts the assumption that $S$ is a solution. Therefore, $S^{\prime}$ is a solution. The solution $S^{\prime}$ contains either $\mathrm{Twins}(u_{1})$ or $\{a_{1}\}$, and therefore Rule (B6) is safe.

The branching vector of Rule (B6) is at least $(1,1)$.

Now suppose that Rule (B4) cannot be applied. We choose non-twin vertices $u_{1},u_{2}\in I_{0}$, a vertex $a\in N(u_{1})\cap N(u_{2})$, and a vertex $v\in I\setminus I_{0}$ that is not adjacent to $a$ as follows.

If the vertices of $I_{0}$ form a sunflower, pick arbitrary non-twin vertices $u_{1},u_{2}\in I_{0}$. Pick $a\in N(u_{1})\cap N(u_{2})$. Since Rule (R4) cannot be applied, there is a vertex $v\in I$ such that $a\notin N(v)$. Since $a\in N(u)$ for every $u\in I_{0}$ (as the vertices of $I_{0}$ form a sunflower) it follows that $v\in I\setminus I_{0}$. 2. 2.

Otherwise, since Rule (B4) cannot be applied, there is a vertex $v\in I\setminus I_{0}$ such that $\bigcup_{u\in I_{0}}N(u)\not\subseteq N(v)$. Pick $a\in(\bigcup_{u\in I_{0}}N(u))\setminus N(v)$. Since the vertices of $I_{0}$ do not form a sunflower, there are non-twin vertices $u_{1},u_{2}\in I_{0}$ such that $a\in N(u_{1})\cap N(u_{2})$.

Since Rule (B4) cannot be applied, there is a vertex $w\in I\setminus I_{0}$ such that, without loss of generality, $a_{1}\notin N(w)$.

(B7)  Branch on $\{u_{1}\}$, $(N(v)\cap N(w))\setminus N(u_{1})$, and on the sets in $\{\{a\},\{v\}\}\circ\{\{a_{1}\},\{w,a_{2}\},\{w,u_{2}\}\}$.

The proof of the safeness of Rule (B7) is similar to the proof for Rule (B3). To bound the branching vector of Rule (B7) we use the following lemma.

Lemma 5**.**

$|(N(v)\cap N(w))\setminus N(u_{1})|\geq 2$.

Proof.

Since $|N(v)|>|N(u_{1})|$, we have that $|N(v)\setminus N(u_{1})|>|N(u_{1})\setminus N(v)|\geq 1$. Similarly, $|N(w)\setminus N(u_{1})|>|N(u_{1})\setminus N(w)|\geq 1$. By Lemma 2 and the fact that $a_{1}\in N(u_{1})\setminus N(w)$ we have that $a\in N(w)$.

We consider two cases. If $|N(v)|>|N(w)|$ then by Lemma 2 and the fact that $a\in N(w)\setminus N(v)$ we have that $N(w)\setminus N(u_{1})\subseteq N(w)\setminus\{a\}\subseteq N(v)$. Therefore, $(N(v)\cap N(w))\setminus N(u_{1})=N(w)\setminus N(u_{1})$ and the lemma follows since $|N(w)\setminus N(u_{1})|\geq 2$.

If $|N(v)|\leq|N(w)|$ then by Lemma 2 and the fact that $a_{1}\in N(v)\setminus N(w)$ we have that $N(v)\setminus N(u_{1})\subseteq N(v)\setminus\{a_{1}\}\subseteq N(w)$. Therefore, $(N(v)\cap N(w))\setminus N(u_{1})=N(v)\setminus N(u_{1})$ and the lemma follows since $|N(v)\setminus N(u_{1})|\geq 2$. ∎

By Lemma 5, the branching vector of Rule (B7) is at least $(1,2,2,2,3,3,3,3)$.

The rule with largest branching number is Rule (B4) and its branching number is at most 2.733. Therefore, the running time of the algorithm is $O^{*}(2.733^{k})$.